# Question Video: Finding the First Derivative of an Exponential Function

Use the integral test to determine whether the series β_(π = 1)^(β) (1/π) converges or diverges.

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### Video Transcript

Use the integral test to determine whether the series, which is the sum from π equals one to β of one over π, converges or diverges.

Letβs start by recalling the integral test. The integral test tells us that if π is a continuous, positive, decreasing function on the interval between π and β and that π of π is equal to π π. Then, if the integral from π to β of π of π₯ with respect to π₯ is convergent, so is the sum from π equals π to β of π π. And secondly, if the integral from π to β of π of π₯ with respect to π₯ is divergent, so is the sum from π equals π to β of π π. Now, in our case, our series is the sum from π equals one to β of one over π. Therefore, π π is equal to one over π. Using the fact that π of π is equal to π π, we can say that π of π₯ is equal to one over π₯.

Now, we need to check that π of π₯ is a continuous, positive, and decreasing function on the interval between π and β. Noting that since our sum goes from π equals one, π must be equal to one. Letβs start by checking the continuity of our function over this interval. The only discontinuity of π of π₯ occurs when π₯ is equal to zero. However, zero is not in our interval between one and β. Therefore, this function must be continuous between one and β. On this interval, π₯ is always positive. Therefore, one over π₯ must also be positive. And so, weβve satisfied that condition too. Now, if π₯ starts at one and gets larger and larger and larger, then one over π₯ will get smaller and smaller and smaller. Therefore, we can say that this is a decreasing function on our interval.

Now that weβve satisfied these three conditions, weβre able to use the integral test. We need to work out whether the integral from one to β of one over π₯ with respect to π₯ is convergent or divergent. Now, we know that the differential of the natural logarithm of π₯ with respect to π₯ is equal to one over π₯. Therefore, the natural logarithm of π₯ is the antiderivative of one over π₯. Hence, we have that our integral is equal to the natural logarithm of π₯ between one and β. When using the bound of β, we need to take the limit as π₯ tends to β of the natural logarithm of π₯. Then, for our lower bound of one, we simply subtract the natural logarithm of one.

Now, letβs evaluate this limit, the natural logarithm as an increasing function. Therefore, as π₯ gets larger, the natural logarithm of π₯ also gets larger. Hence, we can say that this limit is also equal to β. We also have that the natural logarithm of one is equal to zero. However, regardless of what constant this term is, this will not affect our answer of β. This tells us that our integral must be divergent. Hence, by the integral test, we can say that the sum from π equals one to β of one over π is divergent.